Sum of the areas of two squares is 468 m². If the difference of their perimeters are 24 m, find the sides of the two squares.

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  Q) Sum of the areas of two squares is 468 m². If the difference of their perimeters are 24 m, find the sides of the two squares. Solution) Let the sides of the two squares be x m and y m. Thus, their perimeters will be 4x and 4y and areas will be x² and y². Hence, according to question – =4x-4y=24 ⇒x-y=6 ⇒x=y+6 And x²+y² = 468 Substituting value of x- (y+6)²+y²=468 ⇒36+y²+12y+y²=468 ⇒2y²+12y-432=0 ⇒y²+6y-216=0 ⇒y²+18y-12y-216=0 ⇒y(y+18)-12(y+18)=0 ⇒(y+18)(y-12)=0 ⇒y=-18 or 12 Since, side cannot be negative. Therefore, the sides of the square are 12 m and (12+6)m i.e 18 m.
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The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

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Q)  The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Solution) Let the larger number be x and smaller number be y 

According to question x² - y² = 180 and y² = 8x 
⇒ x² - 8x = 180 x² - 8x - 180-0 
⇒ x² - 18x + 10x - 180 = 0 
⇒ x ( x - 18 ) + 10 ( x - 18 ) = 0 
⇒ ( x - 18 ) ( x + 10 ) = 0 
⇒x = 18 , -10 

Since , larger cannot be negative as 8 times of the larger number will be negative and hence , the square of the smaller number will be negative which is not possible . 

Therefore , the larger number will be 18 . .. y² = 8 ( 18 ) ⇒ y² = 144 y = ± 12
 Hence , smaller number be ± 12 . 
Therefore , the numbers are 18 and 12 or 18 and - 12 .


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